The figure shows the orbit of a planet P arou | vedclass

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(C) According to the Virial Theorem for a system bound by an inverse-square law force (like gravity),the time-averaged kinetic energy $\langle K \rang

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Both $D$ and $C$

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(C) According to the Virial Theorem for a system bound by an inverse-square law force (like gravity),the time-averaged kinetic energy $\langle K \rangle$ and potential energy $\langle U \rangle$ are related by $\langle U \rangle = -2 \langle K \rangle.$ For a planet in an elliptical orbit,the total mechanical energy $E$ is constant and is given by $E = K + U = -\frac{GMm}{2a},$ where $a$ is the semi-major axis. At any point in the orbit,the potential energy is $U = -\frac{GMm}{r}.$ Since the total energy $E = K + U = -\frac{GMm}{2a},$ we have $K = E - U = -\frac{GMm}{2a} - (-\frac{GMm}{r}) = GMm(\frac{1}{r} - \frac{1}{2a}).$ Comparing the magnitudes $|U| = \frac{GMm}{r}$ and $|K| = GMm|\frac{1}{r} - \frac{1}{2a}|,$ we observe that $|U| > |K|$ holds true for all points in the orbit,including $C$ and $D.$ Specifically,at any point $r$ in the orbit,since $r$ is always less than $2a$ (the major axis length),the condition $|U| > |K|$ is satisfied throughout the entire elliptical path.

Column $-I$	Column $-II$
$(A)$ Kinetic energy	$(p)$ $-\frac{GM_Em}{2r}$
$(B)$ Potential energy	$(q)$ $\sqrt{\frac{GM_E}{r}}$
$(C)$ Total energy	$(r)$ $-\frac{GM_Em}{r}$
$(D)$ Orbital velocity	$(s)$ $\frac{GM_Em}{2r}$

$(a)$ Gravitational constant $(G)$	$(i)$ $[L^{2}T^{-2}]$
$(b)$ Gravitational potential energy	$(ii)$ $[M^{-1}L^{3}T^{-2}]$
$(c)$ Gravitational potential	$(iii)$ $[LT^{-2}]$
$(d)$ Gravitational intensity	$(iv)$ $[ML^{2}T^{-2}]$

The figure shows the orbit of a planet $P$ around the sun $S.$ $AB$ and $CD$ are the minor and major axes of the ellipse,respectively. If $U$ is the potential energy and $K$ is the kinetic energy of the planet,then $|U| > |K|$ at:

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